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mariarad [96]
3 years ago
7

What happens when any ray of light strikes a concave mirror, after passing through its focal point?

Physics
2 answers:
nekit [7.7K]3 years ago
6 0

Answer:

Option (D)

Explanation:

The rules to draw the ray diagrams for concave mirror are given below:

1. When a ray of light incident on the pole of concave mirror at an angle of incidence, then after reflection, the angle of incidence is equal to the angle of reflection.

2. When incident ray of light passes through the centre of curvature of the mirror, then the reflected ray retraces its path.

3. When the incident ray of light passing through the focus of the concave mirror, then after reflection the reflected ray becomes parallel to the principal axis and vice versa.

baherus [9]3 years ago
5 0
D) The reflected ray travels parallel to the principal axis
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Explanation:

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3 years ago
What conclusions can you make if a hockey goalie fails to block pucks shot in the upper right hand corner of the net
Lemur [1.5K]

Given what we know, we can confirm that this result from the goalie is a clear indicator of room for improvement in the reaction speed and visual coordination for this area of the net.

<h3>How can the goalie improve reaction speeds to this area?</h3>

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3 0
2 years ago
Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,
Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the  the second force(F₂)

F₂x= −19.0 N

F₂y=   16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }

F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }

F_{n} = 30.11N

Direction of the net force (β)

\beta =tan^{-1} (\frac{29.13}{7.62} )

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

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